In a regular Cartesian coordinate system find the coordinates of six lattice points
that are the vertices of a regular hexagon (being a lattice point means that the
point has integer coordinates).

Solution: One has to be careful in how one interprets this question
(and I had to be careful in how I worded it). One might give the following
"proof" that the desired configuration in impossible. If it were possible
we could then take the average of the six points (or of just two opposite
points) to find the center of the hexagon. Then, together with any two
adjacent vertices, we would have an equilateral triangle with all three points
having rational coordinates. But this is impossible - we may assume that
one vertex of the equilateral triangle is at (0,0) and one is at (a,b).
Then to get to the third vertex you could start at (a/2, b/2) and go
perpendicularly to this side a distance equal to root(3)/2 of the side length.
Alternatively we could get to the third vertex by rotating the point (a,b) 60
degrees in either direction. Either way, we end up with the third point
being

,
which is obviously not a point with rational coordinates.

To paraphrase Mr. Spock in
the Wrath of Khan this solution shows some innate talent, however it
betrays a tendency for 2 dimensional thinking. Once one gets over this
tendency, many solutions become possible. My favorite is to pick your six
points to be the six permutations of (1,2,3). I.e. One solution is the
hexagon with vertices located at (1,2,3), (1,3,2), (2,3,1), (3,2,1), (3,1,2),
and (3,2,1). If one is just specifying the vertices then the order chosen
is irrelevant. However, the order I picked should help one verify the
validity of the claim since I listed the points in adjacent order. One can
check that each successive pair are root (2) apart from each other and that the
angle determined by any three of these in order is 120 degrees.